LectureNotes6G - Math 598 Feb 11, 2005 1 Geometry and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 598 Feb 11, 2005 1 Geometry and Topology II Spring 2005, PSU Lecture Notes 6 2.5 The inverse function theorem Recall that if f : M N is a diffeomorphism, then df p is nonsingular at all p M (by the chain rule and the observation that f f- 1 is the identity function on M ). The main aim of this section is to prove a converse of this phenomenon: Theorem 1 (The Inverse Function Theorem) . Let f : M N be a smooth map, and dim( M ) = dim( N ) . Suppose that df p is nonsingular at some p M . Then f is a local diffeomorphism at p , i.e., there exists an open neighborhood U of p such that 1. f is one-to-one on U . 2. f ( U ) is open in N . 3. f- 1 : f ( U ) U is smooth. In particular, d ( f- 1 ) f ( p ) = ( df p )- 1 . A simple fact which is applied a number of times in the proof of the above theorem is Lemma 2. Let f : M N , and g : N L be diffeomorphisms, and set h := g f . If any two of the mappings f , g , h are diffeomorphisms, then so is the third. In particular, the above lemma implies Proposition 3. If Theorem 1 is true in the case of M = R n = N , then, it is true in general. 1 Last revised: September 30, 2009 1 Proof. Suppose that Theorem 1 is true in the case that M = R n = N , and let f : M N be a smooth map with df p nonsingular at some p M . By definition, there exist local charts ( U, ) of M and ( V, ) of N , centered at p and f ( p ) respectively, such that f := - 1 f is smooth. Since and are diffeomorphisms, d p and d f ( p ) are nonsingular. Consequently, by the chain rule, d f o is nonsingular, and is thus a local diffeomorphism. More explicitly, there exists open neighborhoods A and B of the origin o of R n such that f : A B is a diffeomorphism. Since : - 1 ( A ) A is also a diffeomorphism, it follows that f : - 1 ( A ) B is a diffeomorphism. But f = f . So f : - 1 ( A ) B is a diffeomorphism. Finally, since : - 1 ( B ) B is a diffeomorphism, it follows, by the above lemma, that f : - 1 ( A ) - 1 ( B ) is a diffeomorphism. So it remains to prove Theorem 1 in the case that M = R n = N . To this end we need the following fact. Recall that a metric space is said to be complete provided that every Cauchy sequence of that space converges. Lemma 4 (The contraction Lemma) . Let ( X,d ) be a complete metric space, and < 1 . Suppose that there exists mapping f : X X such that d ( f ( x 1 ) , ( x 2 )) d ( x 1 ,x 2 ) , for all x 1 , x 2 X . Then there exists a unique point x X such that f ( x ) = x ....
View Full Document

This note was uploaded on 08/25/2011 for the course MATH 6456 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.

Page1 / 8

LectureNotes6G - Math 598 Feb 11, 2005 1 Geometry and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online